[drunknmunky]

a uniform, narrow-bored tube is closed at one end, contains some dry air which is sealed by a thread of mercury 15.0cm long.

when the tube is vertical, with the closed end at the bottom,the air column is 20cm long, but when held horizontally, the air column is 24cm long.

calculate the atmospheric pressure.

thx

[Morbo]

Assuming temperature of the gas stays constant:
p₁V₁ = p₂V₂

Or in the case of a cylinder of constant area:
p₁L₁ = p₂L₂

Now, the mercury has the effect of increasing the pressure on the cylinder of air when it is vertical:
p₂ = p₁ + ρgL_mercury

where ρ = the density of mercury

So subbing this in to the first equation:
p₁L₁ = (p₁ + ρgL_mercury)L₂
p₁(L₁-L₂) = (ρgL_mercury)L₂
p₁ = (ρgL_mercury)L₂/(L₁-L₂)
p₁ = (ρgL_mercury)/([L₁/L₂]-1)

Sub in the numbers (I think the density of mercury is 5430 kgm^-3)

p₁ = (5430 kgm^-3 x 9.81ms^-2 x 0.15m)/(0.24m/0.20m - 1)
= 39 951 Nm^-2
= 40 kPa or about 0.4 atm

[teachercol]

Er - mercury is a bit denser than that. Often atmospheric pressure is caluclated in cm of Hg so if you work in units of pressure of cmHg then the problem is easy.

P x 24 = ( P + 15 ) x 20

4 P = 300

P = 75 cm Hg

(normal atm is 76cm hg )

[Morbo]

Crikey. Google failed me. It gave me the density of the planet Mercury

So my answer is (13 500/5430) times bigger = 2.5 * 40 = 100kPa

[teachercol]

Nice one

[Arr10]

teachercol im in love with you for solving this :P thanks